# Need help finishing my Lab report for chem

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#1
Help with lab report
Need Help finishing my lab report.KNOWN VALUES ARE V OF HCL=50ML,M=2.188.V OF NAOH=55.3ML,M=2.088.INTIAL TEMP FOR HCL=22.5C,NAOH IS 23C.this is for part A: Mass of final NACL solution assuming that the density of a 1 M Nacl solution is 1.04 g/ml? Heat actually produced from the HCL-NAOH reaction assuming that the heat capacity of a 1 M Nacl solution is 3.93 J/G.C? Moles of water formed in the reaction? Total theoretical heat expected to be produced by forming the number of moles of water assuming that 57,320 J are produced per mole of water formed? Heat absorbed by the calorimeter(heat expected-heat actually produced)? Heat capacity of the calorimeter(heat absorbed by the calorimeter per degree of temperature change)? For Part B: Total mass of the final solution by adding the mass of intial HCL solution(density of 2 M HCL is 1.03 g/ml)the mass of water (D=1.00g/ml)and the mass of magnesium? Heat required to raise MgCl_2 solution to the max temp assuming that heat capacity of your final mgcl2 solution is 3.97 J/G.C? Heat required to raise the calorimeter to the max temp using the value for the heat capacity of the calorimeter as determined in part A of this experiment? Total heat evolved by the reaction by adding the heats required to raise both the solution and the calorimeter to the max temperature? heat evolved per mole of mg? change in enthalpy for the following specific reaction Mg(s)+2H(aq)>>Mg2+(aq)+H2(g)

i really appreciate if someone could help me out finish this!
0
9 years ago
#2
(Original post by randyx)
Help with lab report
Well, what have you managed so far? I don't think you'll find anybody who's willing to answer a dozen questions for you, especially without evidence of some effort on your part.
1
#3
Part A:

1-Mass of final Nacl Solution=55.3ml+50ml=105.3ml(Volume),Multiply it by given Density to get The mass.(1.04 g/ml)(105.3ml)=110g
2-Heat Produced from the HCL-NAOH reaction assuming heat capacity of 1 M Nacl solution is 3.93 J/G.C= Q=MC/\T (110g)(3.93 j/G.C)(8.5C)=3.7*10^4 J
3-Moles of water formed in the reaction>>STUCK ON THIS ONE
4-Total Theoretical heat expected>>>57320J/X Moles on the above question>>> xxxxJ/Mol
6-Heat Capacity of the Calorimeter>>Q//\T Answer5/8.5C>>xxxxJ

Part B:

I finished this part!
0
9 years ago
#4
Are you sure that your Part A - 1 is correct? 110g doesn't even seem relevant if you take the law of conservation of mass into consideration. I am also working on this lab, but I am stuck on the calculations. Thank you for posting this information, it helps, but are u sure about Part A -1?

Thanks!
0
#5
(Original post by Codyy)
Are you sure that your Part A - 1 is correct? 110g doesn't even seem relevant if you take the law of conservation of mass into consideration. I am also working on this lab, but I am stuck on the calculations. Thank you for posting this information, it helps, but are u sure about Part A -1?

Thanks!
yeh.i double checked it with my TA and she said it was right.
0
9 years ago
#6
Can you help me with part B on question 4 ithink i might have done something wrong before?
1
5 years ago
#7
(Original post by randyx)
Part A:

1-Mass of final Nacl Solution=55.3ml+50ml=105.3ml(Volume),Multiply it by given Density to get The mass.(1.04 g/ml)(105.3ml)=110g
2-Heat Produced from the HCL-NAOH reaction assuming heat capacity of 1 M Nacl solution is 3.93 J/G.C= Q=MC/\T (110g)(3.93 j/G.C)(8.5C)=3.7*10^4 J
3-Moles of water formed in the reaction>>STUCK ON THIS ONE
4-Total Theoretical heat expected>>>57320J/X Moles on the above question>>> xxxxJ/Mol
6-Heat Capacity of the Calorimeter>>Q//\T Answer5/8.5C>>xxxxJ

Part B:

I finished this part!
0
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